39. A metallic wire of electrical resistance 40 Ω is bent in the form of a square loop. The resistance between any two diagonally opposite corners is Ω.
Resistance Between Diagonally Opposite Corners of a Square Wire Loop
Understanding the Square Wire Resistance Problem
This question is based on the distribution of electrical resistance in a uniform metallic wire and the combination of resistances in series and parallel. A metallic wire having a total resistance of 40 Ω is bent into the shape of a square. We need to calculate the equivalent resistance measured between any two diagonally opposite corners of the square.
The important point is that the original wire is uniform. Therefore, its resistance is distributed uniformly along its entire length. When the wire is bent into a square, its total length is divided into four equal sides. As resistance is directly proportional to the length of a uniform wire, each side of the square has one-fourth of the total resistance.
Step 1: Calculate the Resistance of Each Side of the Square
The total resistance of the metallic wire is:
R = 40 Ω
The wire is bent into a square containing four equal sides. Therefore, the resistance of each side is:
Resistance of each side = Total resistance/4
Resistance of each side = 40/4
Resistance of each side = 10 Ω
Thus, every side of the square loop has an electrical resistance of 10 Ω.
Step 2: Identify the Two Current Paths Between Diagonally Opposite Corners
Consider two diagonally opposite corners of the square as the terminals between which the equivalent resistance is to be calculated. From one diagonal corner to the other, electric current can travel through two separate paths around the square.
The first path consists of two adjacent sides of the square. Since each side has a resistance of 10 Ω and the two sides are connected one after another along this path, their resistances are in series.
Therefore, the resistance of the first path is:
R1 = 10 Ω + 10 Ω
R1 = 20 Ω
The second path also consists of the remaining two adjacent sides of the square. These two resistances are also connected in series along the second route.
Therefore:
R2 = 10 Ω + 10 Ω
R2 = 20 Ω
Hence, between the two diagonally opposite corners, there are two separate conducting paths, and each path has a resistance of 20 Ω.
Step 3: Recognize the Parallel Combination
Both 20 Ω paths begin at the same corner and end at the same diagonally opposite corner. Therefore, these two paths are connected in parallel.
For two resistances R1 and R2 connected in parallel, the equivalent resistance is given by:
1/Req = 1/R1 + 1/R2
Substituting R1 = 20 Ω and R2 = 20 Ω:
1/Req = 1/20 + 1/20
1/Req = 2/20
1/Req = 1/10
Therefore:
Req = 10 Ω
Why Are the Two 20 Ω Paths in Parallel?
Two resistive branches are said to be connected in parallel when both branches are connected between the same pair of electrical points. In this square loop, one current path goes around two sides of the square in one direction, while the other current path goes around the remaining two sides in the opposite direction.
Both paths start from the first diagonal corner and reach the same opposite diagonal corner. Therefore, the potential difference across both 20 Ω paths is identical. This is the defining condition for a parallel combination.
As a result, the equivalent resistance is smaller than the resistance of either individual path. Two equal resistances of 20 Ω connected in parallel give an equivalent resistance of 10 Ω.
Alternative Short Method
The result can also be obtained quickly by observing the symmetry of the square loop. The total resistance of the wire is 40 Ω. Two diagonally opposite corners divide the complete wire into two equal-length paths.
Since the original wire is uniform, each half of the wire has half of the total resistance:
Resistance of each half = 40/2 = 20 Ω
These two equal halves are connected between the same pair of diagonal corners and are therefore in parallel. The equivalent resistance of two equal resistances R connected in parallel is R/2.
Hence:
Req = 20/2
Req = 10 Ω
This provides a faster way to solve the problem while giving exactly the same answer.
General Formula for a Uniform Wire Forming a Square Loop
Suppose a uniform wire of total resistance R is bent into a square. Each side then has a resistance of R/4.
Between two diagonally opposite corners, each route contains two sides in series. Therefore, the resistance of each route is:
R/4 + R/4 = R/2
The two equal resistances R/2 are connected in parallel. Therefore:
Req = (R/2)/2
Req = R/4
Thus, for any uniform wire bent into a square loop, the equivalent resistance between diagonally opposite corners is one-fourth of the total resistance of the original wire.
For R = 40 Ω:
Req = 40/4 = 10 Ω
Current Distribution Between the Two Paths
The two paths between the diagonally opposite corners have exactly the same resistance of 20 Ω. Therefore, if a voltage source is connected across these corners, the total current divides equally between the two paths.
This equal current distribution is a direct consequence of the symmetry of the square. Since both branches have the same resistance and experience the same potential difference, the current flowing through each branch must be equal according to Ohm’s law.
The symmetry of the circuit is therefore not only useful for calculating the equivalent resistance but also for understanding how current flows through the square wire loop.
Why the Total Resistance Is Not Simply 40 Ω
The original wire has a resistance of 40 Ω from one end to the other when current must travel through its complete length. After the wire is bent into a closed square loop and the terminals are connected at diagonally opposite corners, current is no longer forced to pass through the entire wire in a single path.
Instead, the current has two separate paths, each containing only half the total wire length. These two paths operate simultaneously and are connected in parallel. Parallel paths reduce the effective resistance, which is why the equivalent resistance between the diagonal corners is only 10 Ω rather than 40 Ω.
Final Answer
Resistance between diagonally opposite corners = 10 Ω
The 40 Ω uniform metallic wire is divided equally among the four sides of the square, so each side has a resistance of 10 Ω. Between two diagonally opposite corners, there are two paths, each containing two 10 Ω sides in series. Each path therefore has a resistance of 20 Ω. These two 20 Ω paths are connected in parallel, giving an equivalent resistance of 10 Ω.
